各位老师，我看见了一个美妙的结论，可我不会证明，谁会
https://tieba.baidu.com/p/7760781063?pn=2 \begin{align*}&b^{10}(a^{2}+b^{2})^{2}x^{6}-a^{2}b^{6}(2a^{6}-13a^{4}b^{2}-b^{6})x^{4}y^{2}-a^{2}b^{10}(a^{2}-b^{2})^{2}x^{4}\\
&\qquad+a^{10}(a^{2}+b^{2})^{2}y^{6}-a^{6}b^{2}(2b^{6}-13a^{2}b^{4}-a^{6})x^{2}y^{4}-a^{10}b^{2}(a^{2}-b^{2})^{2}y^{4}\\
&\qquad\qquad+2a^{6}b^{6}(a^{2}-b^{2})^{2}x^{2}y^{2}=0
\end{align*} 葡萄糖 发表于 2022-5-10 22:15
\begin{align*}
&b^{10}(a^{2}+b^{2})^{2}x^{6}-a^{2}b^{6}(2a^{6}-13a^{4}b^{2}-b^{6})x^{4}y^{2}-a^{2}b ...
老师好，可否写一下你的推导过程 \[\left(1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\right)\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{b^{2}x^{2}}{a^{4}}-\frac{a^{2}y^{2}}{b^{4}}\right)^{2}=\left(\frac{2bx^{2}}{a^{3}}+\frac{2ay^{2}}{b^{3}}\right)^{2}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)\]
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